%\magnification=1200
%\baselineskip=18pt

\centerline{ \bf A domain-wall like algorithm for the overlap Dirac operator}
\medskip

\hbox{\bf NOTE: In the SZIN code this method corresponds to OVERLAP\_5D}

\medskip

We are interested in solving the massive hermitian overlap equation
$$
H_o\psi = {1\over 2}\left((1+\mu)\gamma_5 + (1-\mu)\epsilon(H)\right)\psi = b
$$
where $\mu$ is the quark mass and $H$ is the hermitian Wilson-Dirac operator.
In the derivations below, we will choose to normalize $\psi$ so that
the $\epsilon(H)$ has no prefactors. We cast the problem into the form
$$
M \psi = [ A + \hat P^\dagger \epsilon \hat P ] \psi = b
$$
where $\hat P$ is some projection operator and $A$ is a matrix.
If $M$ is the hermitian overlap Dirac opertor, then
$$M=H_o;\ \ \ \  A= {1+\mu\over 1-\mu} \gamma_5 + 
\sum_{i=1}^m {\rm sgn}{\lambda^w_i} \chi_i \chi_i^\dagger$$
$$
\hat P = P_w;\ \ \ \  P_w= 1-\sum_{i=1}^m \chi_i \chi_i^\dagger
$$
where
$$H_w \chi_i = \lambda^w_i \chi_i$$
where $\lambda^w_i$ are the low lying eigenvalues of the hermitian
Wilson-Dirac operator.
We note that $\psi$ has been rescaled $\psi\rightarrow {{2}\over{1-\mu}}\psi$.
If $M$ is the square of the hermitain overlap Dirac operator, then
$$M=H_o^2; \ \ \ \ \  A = \pm {1+ \mu^2\over 1-\mu^2} 
+ P_\pm \sum_{i=1}^m {\rm sgn}{\lambda^w_i} \chi_i \chi_i^\dagger P_\pm;$$
$$
\hat P = P_w P_\pm; \ \ \ \
P_\pm = {1\over 2} (1 \pm \gamma_5); \ \ \ \ \gamma_5 b = \pm b $$
In writing the above equations, we have removed some overall factors
that only depend on $\mu$. 
 
We start with the assumption that we have a rational function 
approximation for $\epsilon(H)$ which we will write as
$P(H)/Q(H)$. We note that $P(H)$ is an odd function of $H$
and $Q(H)$ is an even function of $H$. In this note we will
convert the problem of solving
$$
(A + \hat P^\dagger {P(H)\over Q(H)} \hat P ) \psi = b,
$$
into an equivalent problem of 
$$D_w \phi = c$$
where $D_w$ has the structure of a five dimensional operator with
a finite extent in the fifth direction,
$\phi$ is a five dimensional vector with one four dimensional
component being $\psi$ and $c$ is a five dimensional source with
$b$ being the only non-zero four dimensional component. 
The essence of the idea is contained in Neuberger's {\bf hep-lat/9901003}.

Let us write
$$
{P(H)\over Q(H)} = \alpha_{2N} H + { \alpha^2_{2N-1} \over 
                        H + {\alpha^2_{2N-2} \over
                        H + {\alpha^2_{2N-3} \over
                        H + {\cdots \over
                   \cdots + {\alpha^2_1 \over
                        H + {\alpha^2_0 \over
                        H   }}}}}}
$$
If we simplify this expression into rational form we get 
$$
{P(H)\over Q(H)}  = { H \sum_{n=0}^N c_n H^{2n} \over \sum_{n=0}^N d_n H^{2n}}
$$
which we can relate to rational polynomial approximations of $\epsilon(H)$.

We will now write down the five dimensional operator. The extent
in the fifth direction is $2N+1$.
$$
\pmatrix{
H & \alpha_0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \cr
\alpha_0 & -H & \alpha_1 & 0 & \cdots & 0 & 0 & 0 & 0 \cr
0 & \alpha_1 & H & \alpha_2 & \cdots & 0 & 0 & 0 & 0 \cr
0 & 0 & \alpha_2 & 0 & \cdots & 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & 0 & \cdots & \alpha_{2N-3} & H & \alpha_{2N-2} & 0 \cr
0 & 0 & 0 & 0 & \cdots & 0 & \alpha_{2N-2} & -H & \alpha_{2N-1}\hat P \cr
0 & 0 & 0 & 0 & \cdots & 0 & 0 & \alpha_{2N-1}\hat P^\dagger & 
A +\alpha_{2N}\hat P^\dagger H \hat P \cr}
\pmatrix{\phi_0\cr \phi_1\cr\phi_2\cr\cdots\cr \phi_{2N-2} \cr \phi_{2N-1}
\cr \psi \cr }
=
\pmatrix{0\cr 0\cr 0\cr\cdots\cr 0 \cr 0
\cr b \cr }
$$

To see the connection, one simply has to solve the above equation for
$\psi$ given $b$ by elminating all the $\phi$ variables. From the
first row, we have
$$\phi_0 = -{\alpha_0\over H} \phi_1.$$
and from the second row we then get
$$\phi_1 = {\alpha_1 \over H + {\alpha_0^2\over H}} \phi_2.$$
Then it is easy to show by recursion that
$$\phi_i = { (-1)^{i-1} \alpha_i \over 
                        H + {\alpha^2_{i-1} \over
                        H + {\alpha^2_{i-2} \over
                        H + {\cdots \over
                   \cdots + {\alpha^2_1 \over
                        H + {\alpha^2_0 \over
                        H   }}}}}} \phi_{i+1}
$$
The last but one row gives us the relation
$$\phi_{2N-1} = { \alpha_{2N-1} \over 
                        H + {\alpha^2_{2N-2} \over
                        H + {\alpha^2_{2N-3} \over
                        H + {\cdots \over
                   \cdots + {\alpha^2_1 \over
                        H + {\alpha^2_0 \over
                        H   }}}}}} \psi
$$
and the last equation gives us the desired equation we wanted to
solve in the first place. This ends our connection.

Because of where the mass term $A$ appears in $D_w$, we can not
use a multi-shift Krylov space solver for handling multiple quark masses.

%\end

Consider a non-hermitian form to $H_o$ which still has a hermitian
solution for $H_o\psi = \chi$.
$$
{P(H)\over Q(H)} = \alpha_{N} H + {\alpha^2_{N-1}H \over 
                        H^2+\beta_{N-1} + {\alpha^2_{N-2} \over
                        H^2+\beta_{N-2} + {\alpha^2_{N-3} \over
                        H^2+\beta_{N-3} + {\cdots \over
                   \cdots + {\alpha^2_1 \over
                        H^2+\beta_1+ {\alpha^2_1 \over
                        H^2+\beta_0}}}}}}
$$
where now ${\rm deg}(P(H))=2N+1$, ${\rm deg}(Q(H)) = 2N$, and $P(H)$
is odd and $Q(H)$ is even. We restrict $N$ to be even.
The resulting five dimensional operator has the form
$$
\pmatrix{
H^2+\beta_0 & \alpha_0 & 0 & \cdots & 0 & 0 & 0 & 0 \cr
\alpha_0 & -H^2-\beta_1 & \alpha_1 & \cdots & 0 & 0 & 0 & 0 \cr
0 & \alpha_1 & H & \cdots & 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & \cdots & 0 & 0 & 0 & 0 \cr
0 & 0 & 0 & \cdots & \alpha_{N-3} & H^2+\beta_{N-2} & \alpha_{N-2} & 0 \cr
0 & 0 & 0 & \cdots & 0 & \alpha_{N-2} & -H^2-\beta_{N-1} & \alpha_{N-1}H\hat P \cr\
0 & 0 & 0 & \cdots & 0 & 0 & \alpha_{N}\hat P^\dagger & 
A + \alpha_{2N}\hat P^\dagger H \hat P\cr}
\pmatrix{\phi_0\cr \phi_1\cr\phi_2\cr\cdots\cr \phi_{N-1} \cr \phi_{N}
\cr \psi \cr }
=
\pmatrix{0\cr 0\cr 0\cr\cdots\cr 0 \cr 0
\cr b \cr }
$$
This operator is not hermitian, but the resulting 4 dimensional
operator for $\psi$ is hermitian. It is also not positive (semi)definite.
This operator requires $2N+2$ applications of $H$, so one more
than the optimal rational polynomial form above; however, it only requires $N$
extra fields instead of $2N$.


\end




